A collision cart runs off the edge of a lab table that is 0.95 meters high. How fast did the collision cart leave the table and hit the floor when the cart lands 0.40 meters from the base of the table?
Answers 0.91 m/s, 4.41 m/s
x = vx*t
y = 0.95 - 1/2*g*t^2
where x is the displacement in the horizontal direction, y is the displacement in the vertical direction, and t is time, vx is the speed of the cart in the x direction, and g is the acceleration due to gravity = 9.8
At it's landing, x = 0.4, y = 0:
0.4 = vx*t
0 = 0.95 - 1/2*9.8*t^2
t = 0.44 s s
vx = 0.91 m/s
Well, if the collision cart left the table and hit the floor, I would say it hit the floor pretty quickly. I mean, let's face it, carts aren't known for their hang time. But let's do some calculations anyway.
To find the speed at which the collision cart left the table, we can use the equation:
vf^2 = vi^2 + 2ad
where vf is the final velocity, vi is the initial velocity, a is the acceleration (which is due to gravity and is approximately 9.8 m/s^2), and d is the distance the cart fell.
Given that the cart fell from a height of 0.95 meters and hit the floor 0.40 meters from the base of the table, we can calculate the distance it fell:
d = 0.95 - 0.40
d = 0.55 meters
Now, plugging in the values into the equation, we get:
vf^2 = vi^2 + 2(9.8)(0.55)
To find the final velocity, we can rearrange the equation:
vf = sqrt(vi^2 + 2ad)
But we don't know the initial velocity, so we need to find it. Well, considering that the cart left the table, we can assume that its initial velocity is zero. So the equation simplifies to:
vf = sqrt(2ad)
Plugging in the values, we get:
vf = sqrt(2(9.8)(0.55))
vf = sqrt(10.78)
vf ā 3.28 m/s
So according to my calculations, the speed at which the collision cart left the table and hit the floor is approximately 3.28 m/s. But hey, who knows, maybe the cart had some extra oomph and hit the floor even faster! After all, it's a collision cart, not a leisurely stroll cart.
To solve this problem, we can use the principles of conservation of energy. The potential energy the cart has at the top of the table is converted into kinetic energy when it hits the floor.
Step 1: Determine the potential energy of the cart at the top of the table.
Potential energy (PE) is given by the formula PE = m * g * h, where m is the mass of the cart, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the table (0.95 m). Let's assume the mass of the cart is 1 kg.
PE = 1 kg * 9.8 m/s^2 * 0.95 m
PE = 9.31 J
Step 2: Determine the kinetic energy of the cart when it hits the floor.
Kinetic energy (KE) is given by the formula KE = (1/2) * m * v^2, where v is the velocity of the cart when it hits the floor.
Step 3: Equate the potential energy with the kinetic energy.
Since energy is conserved, we can set the potential energy equal to the kinetic energy.
PE = KE
9.31 J = (1/2) * 1 kg * v^2
9.31 J = 0.5 kg * v^2
Step 4: Solve for the velocity of the cart.
Rearrange the equation and solve for v.
v^2 = (9.31 J) / (0.5 kg)
v^2 = 18.62 J/kg
v = ā(18.62 J/kg)
v ā 4.31 m/s
Step 5: Determine the horizontal velocity of the cart.
The horizontal velocity of the cart does not change during free fall. It remains constant at the point of leaving the table and hitting the floor.
Horizontal velocity = ? (not provided in the question)
Step 6: Use the horizontal velocity and the distance from the base of the table to determine the time it takes for the cart to hit the floor.
Time (t) can be calculated using the formula t = d / v, where d is the distance from the base of the table (0.40 m) and v is the horizontal velocity.
t = 0.40 m / ? (not provided in the question)
Since the horizontal velocity is not provided, we cannot accurately determine the time (and therefore the vertical velocity when the cart hits the floor).
Therefore, the speed at which the cart leaves the table and hits the floor cannot be determined with the given information.
To find the speed at which the collision cart leaves the table and hits the floor, we can use the principles of kinematics. There are a few steps involved in finding the solution:
1. First, we need to determine the time it takes for the cart to fall from the table to the floor. We can use the equation for the time of flight in free fall:
š” = ā(2š/š)
where š” is the time of flight, š is the vertical distance (height) from the table to the floor, and š is the acceleration due to gravity (approximately 9.8 m/sĀ²).
Substituting the given values, we have:
š” = ā(2 Ć 0.95 m / 9.8 m/sĀ²) ā 0.44 s
2. Next, we need to find the horizontal velocity of the cart as it leaves the table. Since there are no horizontal forces acting on the cart, its horizontal velocity remains constant. Therefore, the horizontal velocity when the cart hits the floor is the same as its horizontal velocity just before it leaves the table.
3. To find the horizontal velocity, we can use the equation for horizontal distance:
š = š£š„ Ć š”
where š is the horizontal distance traveled, š£š„ is the horizontal velocity, and š” is the time of flight.
Rearranging the equation, we have:
š£š„ = š / š”
Substituting the given values, we have:
š£š„ = 0.4 m / 0.44 s ā 0.91 m/s
Therefore, the horizontal velocity of the cart just before it hits the floor is approximately 0.91 m/s.
4. Finally, to find the total speed at which the cart hits the floor, we need to calculate the resultant velocity. This can be found using the Pythagorean theorem:
š£(š”šš”šš) = ā(š£š„Ā² + š£š¦Ā²)
where š£(š”šš”šš) is the total velocity, š£š„ is the horizontal velocity, and š£š¦ is the vertical velocity.
Since the cart is only affected by gravity in the vertical direction, the vertical velocity can be found using the equation:
š£š¦ = š Ć š”
Substituting the given values, we have:
š£š¦ = 9.8 m/sĀ² Ć 0.44 s ā 4.31 m/s
Now we can calculate the total velocity:
š£(š”šš”šš) = ā(0.91 m/s)Ā² + (4.31 m/s)Ā² ā 4.41 m/s
Therefore, the total speed at which the collision cart hits the floor is approximately 4.41 m/s.